Question

Show that Theorem \(2.36\) and its Corollary become false (in \(R^{1}\), for example) if the word "compact" is replaced by "closed" or by "bounded."

Short answer

Question: Provide counterexamples to show that Theorem 2.36 and its Corollary are false in the case of R^1 when "compact" is replaced by "closed" or by "bounded." Answer: For closed sets, a counterexample for Theorem 2.36 is the sets A = {1, 2, 3, ...} and B = {0}, where their distance is 1 instead of the infimum of distances. For the Corollary, a counterexample is the closed sets A = {0} and B = {1, 1/2, 1/3, ...}, which have an infimum distance of 0, not a positive number. For bounded sets, a counterexample for Theorem 2.36 is the sets A = {1/n : n ∈ N} and B = {1 + 1/n : n ∈ N}, where both are bounded but not closed, and Theorem 2.36 does not hold true. For the Corollary, a counterexample is the disjoint bounded sets A = {1/n : n ∈ N} and B = {2 + 1/n : n ∈ N}, which have an infimum distance that is not positive.

Step-by-step solution

Counterexample for Theorem 2.36 with closed sets

Consider the following two closed sets in R^1: A = {1, 2, 3, ...} and B = {0}. Here, A and B are both closed since they contain all their limit points. However, the distance between them is 0+1=1, which is not the infimum of distances between all pairs. So in this case, Theorem 2.36 does not hold true.

Counterexample for Corollary with closed sets

Consider the closed sets A = {0} and B = {1, 1/2, 1/3, ...}. Even though they are closed and disjoint, they do not have a positive distance between them because the infimum of distances is 0, not a positive number. Therefore, the Corollary does not hold true for closed sets either. #Case 2: Bounded Sets#

Counterexample for Theorem 2.36 with bounded sets

Consider the following two bounded sets in R^1: A = {1/n : n ∈ N} and B = {1 + 1/n : n ∈ N}. Both A and B are bounded (since they are contained within an interval [-1, 2] for example), but neither is closed. The infimum of the distances between these pairs is 1, however, since the sets are not compact, Theorem 2.36 does not hold true.

Counterexample for Corollary with bounded sets

Now consider two disjoint bounded sets in R^1: A = {1/n : n ∈ N} and B = {2 + 1/n : n ∈ N}. Though they are bounded and disjoint, the distance between A and B is not positive, since the sequences in A and B approach 0 and 2 respectively, and hence the distance between their limit points is 2. The infimum of distances between A and B is not positive, so the Corollary does not hold true for bounded sets.

Key concepts

Compact Sets in R^1

In the realm of real analysis, a set in the real numbers \( R^1 \) is called compact if it is both closed and bounded. This definition encapsulates the intuitive notion that a compact set is limited in extent and contains all its limit points (the points that can be 'approached' by the elements of the set).
The power of compactness as a concept lies in its ability to guarantee the presence of certain important properties. One such instance is illustrated by Theorem 2.36, which often deals with the behavior of functions or sequences on compact sets. The essence of this theorem is that it provides conditions under which certain desirable outcomes (like the existence of maximum or minimum values) are assured.
To provide an intuitive understanding, consider the visual image of a closed loop or an interval with both ends included. Such shapes in \( R^1 \) are compact since they have clear boundaries and do not stretch out indefinitely. Conversely, when Theorem 2.36 is applied to sets that are not compact, as seen in the exercise, the expected results break down. Thus, highlighting the crucial nature of compactness in mathematical analysis.

Closed Sets in Mathematical Analysis

The notion of closed sets is pivotal in the study of mathematical analysis. A set is termed closed if it includes all its boundary or limit points. In simpler terms, a closed set does not allow for 'leakage;' every point that is infinitely close to the set is actually part of the set.
For example, the set of all integers \( \{...,-2,-1,0,1,2,...\} \) is closed because there are no 'gaps' between the integers where a non-integer point could slip in and still be infinitely close to the integers. However, the exercise shows that being closed is not enough to satisfy Theorem 2.36. This is because closedness does not control how far the set stretches out. If a closed set is also unbounded, it can extend indefinitely, leading to counterexamples where the conclusions of Theorem 2.36 fail to hold, as demonstrated in the provided closed sets' examples.

Bounded Sets in Real Analysis

The term bounded in real analysis signifies that a set is contained within a certain range, with a well-defined 'boundary' beyond which the set's elements cannot be found. Formally, a set \( A \) in \( R^1 \) is said to be bounded if there exists a real number \( M > 0 \) such that for every element \( a \) in \( A, |a| \leq M \) holds true.
An easy way to visualize boundedness is to think of a fence around a field, keeping the sheep from wandering off indefinitely; within mathematics, the 'fence' is a numerical boundary. Nonetheless, as the exercise demonstrates, a set being bounded does not inherently imply that it will comply with Theorem 2.36 or its Corollary. A bounded set can still approach a point as closely as desired without ever including that point, thereby not satisfying the condition of compactness required for the application of Theorem 2.36.